Difference between revisions of "2014 AMC 8 Problems/Problem 24"

Revision as of 09:21, 27 April 2022

Problem

One day the Beverage Barn sold $252$ cans of soda to $100$ customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?

$\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5$

Video Solution for Problems 21-25

https://www.youtube.com/watch?v=qMb_GCP0mbw

Video Solution

https://www.youtube.com/watch?v=FE0u3Y1FCGk

https://youtu.be/VlJzQ-ZNmmk ~savannahsolver

Solution

In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are $100$ people, the median will be the average of the $50\text{th}$ and $51\text{st}$ largest amount of cans per person. To minimize the first $49$ , they would each have one can. Subtracting these $49$ cans from the $252$ cans gives us $203$ cans left to divide among $51$ people. Taking $\frac{203}{51}$ gives us $3$ and a remainder of $50$ . Seeing this, the largest number of cans the $50$ th person could have is $3$ , which leaves $4$ to the rest of the people. The average of $3$ and $4$ is $3.5$ . Thus our answer is $\boxed{\text{(C) }3.5}$

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-A straight one-mile stretch of highway, <math>40</math> feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at <math>5</math> miles per hour, how many hours will it take to cover the one-mile stretch?
+==Problem==
+One day the Beverage Barn sold <math>252</math> cans of soda to <math>100</math> customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?
-Note: <math>1</math> mile= <math>5280</math> feet
+<math>\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5</math>
-<asy>
+==Video Solution for Problems 21-25==
-size(10cm); pathpen=black; pointpen=black;
+https://www.youtube.com/watch?v=qMb_GCP0mbw
-D(arc((-2,0),1,300,360));
-D(arc((0,0),1,0,180));
+==Video Solution==
-D(arc((2,0),1,180,360));
+https://www.youtube.com/watch?v=FE0u3Y1FCGk
-D(arc((4,0),1,0,180));
-D(arc((6,0),1,180,240));
+https://youtu.be/VlJzQ-ZNmmk ~savannahsolver
-D((-1.5,1)--(5.5,1));
-D((-1.5,0)--(5.5,0),dashed);
+==Solution==
-D((-1.5,-1)--(5.5,-1));
+In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are <math>100</math> people, the median will be the average of the <math>50\text{th}</math> and <math>51\text{st}</math> largest amount of cans per person. To minimize the first <math>49</math>, they would each have one can. Subtracting these <math>49</math> cans from the <math>252</math> cans gives us <math>203</math> cans left to divide among <math>51</math> people. Taking <math>\frac{203}{51}</math> gives us <math>3</math> and a remainder of <math>50</math>. Seeing this, the largest number of cans the <math>50</math>th person could have is <math>3</math>, which leaves <math>4</math> to the rest of the people. The average of <math>3</math> and <math>4</math> is <math>3.5</math>. Thus our answer is <math>\boxed{\text{(C) }3.5}</math>
-</asy>
-<math> \textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3} </math>
 ==See Also==
 {{AMC8 box|year=2014|num-b=23|num-a=25}}
 {{MAA Notice}}

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